Experimental materials

Compared with hardwood, the anatomical structure of softwood is simpler and the variability is less obvious. Therefore, in this study Chinese fir (*Cunninghamia lanceolata* (Lamb.) Hook.) was utilized to compare the two analytical methods. Chinese fir was sourced from Hebei Province, China. The samples with the size of 5mm(R)×5mm(L)×1mm(T) were cut from the green sapwood with straight texture and no defect.

The samples were placed in a desiccator in which the relative humidity was kept at 100% by distilled water at 25°C. A thermometer was used to detect the relative humidity and temperature. The samples were weighed every day. When the mass change of the samples between two weights was less than 0.02%, the samples were deemed to have reached moisture equilibrium (Li and Ma 2021).

Pore size distribution calculation

Two pore size distribution calculation methods were used and compared in this study.

*The continuous method* proposed in this paper solves the endothermic differential equation of the freeze-thaw process of water in the wood to obtain the frozen water content of the samples at different temperatures, and then calculate the pore size distribution. During the heating process of the samples in the low-temperature state, the heat absorbed by the samples is divided into two parts: sensible heat and latent heat. Sensible heat is the heat required to increase the temperature of the substance, including the substance of wood cell wall, the frozen water in the pores, and the unfrozen water. Note that the air in the sample was not included in this study. Latent heat is the heat required for the phase change of frozen water in pores of different sizes. Based on the above, the heat absorption can be expressed by Eq. (2):

$${\Delta }H={m}_{cw}{c}_{cw}{\Delta }T+{m}_{fw}{c}_{i}{\Delta }T+\left({m}_{w}-{m}_{fw}\right){c}_{w}{\Delta }T-{H}_{f}{\Delta }{m}_{fw}$$

2

where \({\Delta }T\) is the difference in temperature (℃); \({\Delta }H\) is the heat absorption corresponding to the temperature change (mJ); \({m}_{cw}\), \({m}_{fw}\), and \({m}_{w}\) are the mass of wood cell wall substance, frozen water, and total water in the samples respectively (mg); \({c}_{cw}\), \({c}_{w}\) and \({c}_{i}\) are the specific heat capacities of wood cell wall, water and ice, respectively (J ℃−1 g−1); \({\Delta }{m}_{fw}\) is the melting amount of frozen water corresponding to temperature changes (mg); \({H}_{f}\) is the specific heat of fusion of pure water (J g−1). Frozen water in wood was assumed to behave like pure water, thus the accepted value of 333.6 J g−1 was used (Repellin and Guyonnet 2005; Dieste et al. 2009).

Differentiate the time on both sides of Eq. (2) to obtain Eq. (3):

$$\frac{dH}{dt}={m}_{cw}{c}_{cw }\lambda +{m}_{fw}\left({c}_{i}-{c}_{w}\right) {\lambda }+{m}_{w}{c}_{w} {\lambda }-{H}_{f}\frac{\text{d}{m}_{fw}}{\text{d}T} {\lambda }$$

3

where \({\lambda }\) is the rate of temperature change, that is, \(\text{d}T/\text{d}t\), and \(t\) represents the time; \(\frac{dH}{dt}\) refers to the time rate of change of heat absorption, that is, heat flow, which can be expressed as a function of temperature; \(\frac{\text{d}{m}_{fw}}{\text{d}T}\) represents the rate of change of frozen water content concerning temperature and it is also a function of temperature. Thus Eq. (4) can be obtained by organizing Eq. (3):

$$\frac{\text{d}{m}_{fw}\left(T\right)}{\text{d}T}-\frac{{c}_{i}-{c}_{w}}{{H}_{f}}{m}_{fw}\left(T\right)=-\frac{1}{{\lambda }{H}_{f}}\frac{\text{d}H}{\text{d}t}\left(T\right)-\frac{{m}_{cw}{c}_{cw}+{m}_{w}{c}_{w}}{{H}_{f}}$$

4

Obviously, Eq. (4) is a non-homogeneous first order linear differential equation for frozen water content \({m}_{fw}\) with respect to temperature \(T\). Solving Eq. (4) can obtain the frozen water content as a function of temperature \({m}_{fw}\left(T\right)\):

$${m}_{fw}\left(T\right)=\frac{1}{{\lambda }{H}_{f}}{e}^{\frac{{c}_{i}-{c}_{w}}{{H}_{f}}T}{\int }_{T}^{0}\frac{\text{d}H}{\text{d}T}\left(T\right){e}^{- \frac{{c}_{i}-{c}_{w}}{{H}_{f}}T}\text{d}T+\frac{{m}_{cw}{c}_{cw}+{m}_{w}{c}_{w}}{{c}_{i}-{c}_{w}}+C{e}^{\frac{{c}_{i}-{c}_{w}}{{H}_{f}}T}$$

5

Since the heat flow function is not clear, the DSC measurement data can be used to calculate the numerical solution of the equation instead of the analytical solution. And it is acceptable that when the temperature is 0°C, there is no frozen water in the wood sample (i.e., when \(T=0\), then \({m}_{fw}\left(T\right)=0\)). Considering this condition into the Eq. (5) can easily obtain the value of the constant \(C\):

$$C=-\frac{{m}_{cw}{c}_{cw}+{m}_{w}{c}_{w}}{{c}_{i}-{c}_{w}}$$

6

There is \({m}_{cw}{c}_{cw}+{m}_{w}{c}_{w}\) in both Eq. (5) and Eq. (6), which just represents the heat capacity of the samples at a specific state when no water is frozen, and thus it can be easily measured at a temperature above 0°C. Finally, the pore size distribution \(\frac{dV}{dD}\) of the samples is derived from the relationship between the content of frozen water and the temperature in Eq. (7).

$$\frac{dV}{dD}=\frac{\left|d{V}_{fw}\right|}{dD}=\frac{d|{m}_{fw}/{\rho }_{fw}|}{dD}=-\frac{1}{{{\rho }}_{fw}}\bullet \frac{d{m}_{fw}}{dT}\bullet \frac{dT}{dD}$$

7

where \({V}_{fw}\) is the volume of frozen water (cm3); \({{\rho }}_{fw}\) is the density of frozen water (g cm−3), the value of 0.9 g cm−3 (the density of ice) was used in this study; \(\frac{d{m}_{fw}}{dT}\) and \(\frac{dT}{dD}\) can be derived from the Eq. (5) and (1), respectively.

*The discontinuous method* was performed as a comparison which took an isothermal step method according to Park et al. (2006b) as well as Zauer et al. (2014b). Pore size distribution was determined by measuring the amount of water that has its melting temperature depressed at each isothermal step. If the frozen water does not melt in the first heating step, the sensible heat can be calculated based on the heat absorption of this step, and the latent heat of melting in the subsequent steps is calculated by subtracting the sensible heat from the total heat:

$${H}_{i}={H}_{t,i}-{c}_{s}{\Delta }T$$

8

$${m}_{fw,i}=\frac{{H}_{i}}{{H}_{f}}$$

9

where \({H}_{i}\) is the latent heat of water at each step (J); \({H}_{t,i}\) is the total heat absorption at each step (J); \({c}_{s}\) is the specific heat capacity of the samples (J ℃−1); \({m}_{fw,i}\) is the mass of frozen water at each step (g).

In summary, there are the following differences between the continuous method and discontinuous method: 1) The continuous method uses a continuous measurement procedure, during which no isothermal stage is required, thus a continuous pore size distribution could be acquired, while the discontinuous method uses an intermittent measurement procedure, and the obtained pore size distribution is discontinuous. Therefore, the former can make better use of the measured data, and greatly reduce the measuring time. 2) The discontinuous method assumes that the specific heat of the samples does not change in the early stage of the experiment, which may cause errors. In the contrast, the continuous method does not require this assumption, thus it is possible to achieve more theoretical accuracy.

DSC measurement

TA Company's DSC-Q2000 instrument was employed for the freeze-thaw test. The samples were cut to a chip with a diameter of about 4mm and put in an aluminum pan for sealing. Each sample was cooled to -30°C and maintained for 5 minutes to ensure that the water in the sample was fully frozen. Due to the different principles of continuous method and discontinuous method as mentioned above, two different DSC heating programs were taken:

Continuous method: The temperature rose to 0°C at a constant rate of 2°C min−1, maintained for 3 minutes, and then continued to rise to 10°C at the same rate. This measurement took 23 minutes in total.

Discontinuous method: The temperature rose step by step to -21, -11.6, -7.3, -5.4, -4.2, -2.8, -2, -1.3 -0.8°C at a rate of 2 ℃ min−1, and maintained to return the heat flow to the baseline in each step, and finally continued to rise to 10°C at the same rate. This process took about 45 minutes. The pore diameters corresponding to different temperatures calculated by the Eq. (1) are shown in Table 1.

Table 1

The pore diameters corresponding to different temperatures.

Temperature (℃) | Pore diameter (nm) |

-21 | 1.9 |

-11.6 | 3.4 |

-7.3 | 5.4 |

-5.4 | 7.3 |

-4.2 | 9.4 |

-2.8 | 14.1 |

-2 | 19.8 |

-1.3 | 30.5 |

-0.8 | 49.5 |

Generally, a slower heating rate and lower sample mass can lead to more accurate enthalpy measurements (Simpson and Barton 1991). However, if the heating rate is too low, the measured signal will show noise and it is difficult to perform the repeatable results. Therefore, a heating rate of 2 ℃ min−1 was used in this study (Zauer et al. 2014b).

After the measurement, the samples were taken out from the aluminum pan, placed in an oven to be dried at 103 ℃ for 8 hours, and the oven-dry mass was weighed, and the moisture content of the samples was calculated.